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A272618
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Irregular array read by rows: n-th row contains (in ascending order) the nondivisors 1 <= k < n such that all the prime divisors p of k also divide n.
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27
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0, 0, 0, 0, 0, 4, 0, 0, 0, 4, 8, 0, 8, 9, 0, 4, 8, 9, 0, 0, 4, 8, 12, 16, 0, 8, 16, 9, 4, 8, 16, 0, 9, 16, 18, 0, 4, 8, 16, 0, 8, 16, 0, 4, 8, 9, 12, 16, 18, 20, 24, 25, 27, 0, 0, 9, 27, 4, 8, 16, 32, 25, 8, 16, 24, 27, 32, 0, 4, 8, 16, 32, 9, 27, 16, 25, 32, 0, 4, 8, 9, 12, 16, 18, 24, 27, 28, 32
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OFFSET
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1,6
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COMMENTS
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The k are the "semidivisors" or nondivisor regular numbers of n as counted by A243822(n).
All nonzero terms k are composite and pertain to composite rows n. This is because prime k must either divide or be coprime to n, and k = 1 is both a divisor of and coprime to n.
Row n for prime p contains zero, since numbers 1 <= k < p must either divide or be coprime to prime p.
Row n for prime powers p^e contains zero, since there is only one prime divisor p of p^e and every power 1 <= m <= e of p divides p^e.
Row n = 4 is a special case of composite n that contains zero. This is because 4 is the smallest composite number; there are no composites k < n.
Thus rows n for composite n > 4 contain at least 1 nonzero value.
In base n, 1/a(n) has a terminating expansion with at least 2 places.
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REFERENCES
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G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, pp. 144-145, Theorem 136.
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LINKS
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EXAMPLE
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For n = 12, the numbers 1 <= k < n such that the prime divisors p of k also divide n are {2, 3, 4, 6, 8, 9}; {2, 3, 4, 6} divide n = 12, thus row n = 12 is {8, 9}.
n: k
1: 0
2: 0
3: 0
4: 0
5: 0
6: 4
7: 0
8: 0
9: 0
10: 4 8
11: 0
12: 8 9
13: 0
14: 4 8
15: 9
16: 0
17: 0
18: 4 8 12 16
19: 0
20: 8 16
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MATHEMATICA
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Table[With[{r = First /@ FactorInteger@ n}, Select[Range@ n,
And[SubsetQ[r, Map[First, FactorInteger@ #]], ! Divisible[n, #]] &]], {n, 30}] /. {} -> 0 // Flatten (* Michael De Vlieger, May 03 2016 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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